Find the last two digits of $7^{100}-3^{100}$ 
Find the last two digits of $7^{100}-3^{100}$

From Euler's theorem one gets that $\phi(100) = 40 \Rightarrow 7^{40} \equiv 1 \pmod{100}, 3^{40} \equiv 1 \pmod{100}.$
I couldn't really work this out without using a calculator to compute the powers. How can I continue from here? I could write the expressions as $7^{100} \equiv 7^{40}\cdot7^{60} \equiv 7^{60} \pmod{100}$, but I would still need to deal with the $7^{60}...$
 A: Alternatively, notice:$$(10-3)^{100}-3^{100} = \sum_{k=0}^{99}\binom{100}{k}10^{100-k}(-3)^k \equiv0 \pmod {100}$$
A: Hint:
The Carmichael function of $100$ is $20$,
so if $\gcd(a,100)=1$ then $a^{20}\equiv1\bmod100$,
so $a^{100}=(a^{20})^5\equiv1\bmod 100$.
A: $\textbf{Hint:}$ Calculate $7^{100}-3^{100}$ separately for modulo $5^2$ and $2^2$.Then,combine them using chinese remainder theorem.It will be easier to calculate I think
A: Yet an other answer, that leads to the result (without using the Euler indicator function). Explicit computations.

*

*Working modulo $4$, i.e. in $\Bbb Z/4$, we have $7^{100}-3^{100}=3^{100}-3^{100}=0$.


*Working modulo $25$, i.e. in $\Bbb Z/25$, note that $1/7=18$ (because of $7\cdot 18=126=1$), so we have $18^{100}(7^{100}-3^{100})=1-54^{100}=1-4^{100}=1-1024^{20}=1-(-1)^{20}=1-1=0$.
So $(7^{100}-3^{100})$ is zero modulo $4$, and modulo $25$, so also modulo $4\cdot 25=100$.
A: For $7$ we have the cycle modulo $100$ with $7, 49, 43, 1$. Then $7^{20}= (7^4)^{5}\equiv 1$.
A: This is my try. As being a high school student, I don't have any idea about $\text{mod}$.
$$7^{100}=(10-3)^{100}=\sum_{r=0}^{100} {{100}\choose{r}}(10)^{100-r}(-3)^r$$ Now, since we want only last two digits, we are interested in terms with index of $10$ less than $2$ as other terms will terminate with unnecessary zeroes.
We've to find last two digits of $${100\choose 99}(10)(-3)^{99}+{100\choose 100}(-3)^{100}-3^{100}\\ =(1000)(-3)^{99}$$
Hence, the last two digits are $00$.
